Quantum Country exercise 10

This is one in a series of posts working through the exercises in the Quantum Country online introduction to quantum computing and related topics. The exercises in the original document are not numbered; I have added my own numbers for convenience in referring to them.

Exercise 10. Show that M \vert e_k \rangle is the kth column of the matrix M and that \langle e_j \vert M \vert e_k \rangle is the jkth entry of M.

Answer: The product of the matrix M and the vector \vert e_k \rangle is a vector, with the jth entry of that vector being the inner product of the jth row of M with \vert e_k \rangle:

\left( M \vert e_k \rangle \right)_j = \sum_l M_{jl}  \vert e_k \rangle_l

We then have

\sum_l M_{jl}  \vert e_k \rangle_l = M_{jk}  \vert e_k \rangle_k + \sum_{l \ne k} M_{jl}  \vert e_k \rangle_l

But by the definition of \vert e_k \rangle we have \vert e_k \rangle_k = 1 and \vert e_k \rangle_l = 0 for l \ne k. So the above reduces to

M_{jk}  \vert e_k \rangle_k + \sum_{l \ne k} M_{jl}  \vert e_k \rangle_l = M_{jk} \cdot 1 + \sum_{l \ne k} M_{lj}  \cdot 0 = M_{jk}

So \left( M \vert e_k \rangle \right)_j = M_{jk} for all j, which means that M \vert e_k \rangle is the kth column of M.

We now consider the product \langle e_j \vert M \vert e_k \rangle for some j. This is the inner product of the vector \langle e_j \vert with the vector \vert M \vert e_k \rangle, which we have just shown is the kth column of M. We therefore have

\langle e_j \vert M \vert e_k \rangle = \sum_l \langle e_j \vert_l M_{lk} = \langle e_j \vert_j M_{jk} + \sum_{l \ne j} \langle e_j \vert_l M_{lk}

But by the definition of \langle e_j \vert we have \langle e_j \vert_j = 1 and \vert e_j \rangle_l = 0 for l \ne j. So the above reduces to

\langle e_j \vert_j M_{jk} + \sum_{l \ne j} \langle e_j \vert_l M_{lk} = 1 \cdot M_{jk} + \sum_{l \ne j} 0 \cdot M_{lk} = M_{jk}

We thus have \langle e_j \vert M \vert e_k \rangle = M_{jk}.

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